In Year 1 you expanded $(a + b)^n$ for a positive whole number $n$, using $^nC_r$ or Pascal's triangle (see The Binomial Expansion). That expansion always stops after $n + 1$ terms. Now $n$ can be negative or a fraction.
1. When $n$ is not a positive integer, no factor $(n - k)$ is ever zero, so the expansion never stops: it is an infinite series.
2. An infinite series only adds up to the right answer if it converges, so the expansion is only valid for $|x| < 1$.
Expand $(1 + x)^{-1}$ in ascending powers of $x$ up to $x^3$, and state the range of validity.
Replace every $x$ in the formula by $bx$, in a bracket. The expansion is valid when $|bx| < 1$, i.e. $|x| < \dfrac{1}{|b|}$.
Expand $\sqrt{1 - 2x}$ up to the term in $x^3$, and state the values of $x$ for which it is valid.
In the expansion of $(1 + ax)^n$ the coefficient of $x$ is $-6$ and the coefficient of $x^2$ is $27$. Find $a$ and $n$, and the coefficient of $x^3$.
The formula needs a $1$ at the front of the bracket. So take out a factor of $a$ first — remembering that it comes out raised to the power $n$.
Find the first four terms of the expansion of $\dfrac{1}{(2 + 3x)^2}$, and state the range of validity.
When $|x|$ is small, the powers $x^2, x^3, \ldots$ are tiny, so the first few terms give an excellent approximation. Choose a value of $x$ that is inside the range of validity and as small as possible.
Use the expansion of $\sqrt{1 - 2x}$ from Worked Example 2, with $x = 0.01$, to find an approximation for $\sqrt{2}$.
Expand each bracket separately, then multiply out, keeping only the powers you need. For a sum of fractions, use partial fractions first.
Expand $\dfrac{1 + x}{1 - 2x}$ up to $x^3$.
(a) Express $\dfrac{4 - 5x}{(1 + x)(2 - x)}$ in partial fractions. (b) Hence expand it up to $x^3$ and state the range of validity.
General expansion
$(1+x)^n = 1 + nx + \tfrac{n(n-1)}{2!}x^2 + \cdots$
Validity
$(1+bx)^n$: valid for $|bx| < 1$.
Positive integer $n$
Finite, valid for all $x$.
$(a+bx)^n$
Take out $a^n$: $a^n(1 + \tfrac{b}{a}x)^n$.
Brackets
$(-2x)^2 = 4x^2$, $(-2x)^3 = -8x^3$.
$(1+x)^{-1}$
$1 - x + x^2 - x^3 + \cdots$
$(1-x)^{-1}$
$1 + x + x^2 + x^3 + \cdots$
Combined
Take the narrowest range of validity.
Approximating
Pick a small $x$ inside the valid range.
Expand $(1 + x)^{-3}$ up to $x^3$.
▶ Show solution
$1 + (-3)x + \dfrac{(-3)(-4)}{2}x^2 + \dfrac{(-3)(-4)(-5)}{6}x^3$
$= 1 - 3x + 6x^2 - 10x^3 + \cdots$, valid for $|x| < 1$.
Expand $\sqrt{1 + 4x}$ up to $x^3$, stating the range of validity.
▶ Show solution
$1 + \tfrac12(4x) + \left(-\tfrac18\right)(4x)^2 + \tfrac{1}{16}(4x)^3$
$= 1 + 2x - 2x^2 + 4x^3 + \cdots$, valid for $|4x| < 1$, i.e. $|x| < \tfrac14$.
Find the first three terms of the expansion of $\dfrac{1}{(2 - x)^2}$ and of the term in $x^3$, and state the range of validity.
▶ Show solution
$(2 - x)^{-2} = \tfrac14\left(1 - \tfrac{x}{2}\right)^{-2}$.
$\left(1 - \tfrac{x}{2}\right)^{-2} = 1 - 2\left(-\tfrac{x}{2}\right) + 3\left(\tfrac{x^2}{4}\right) - 4\left(-\tfrac{x^3}{8}\right) = 1 + x + \tfrac34x^2 + \tfrac12x^3$
So $\dfrac{1}{(2-x)^2} = \tfrac14 + \tfrac14x + \tfrac{3}{16}x^2 + \tfrac18x^3 + \cdots$, valid for $|x| < 2$.
State, with a reason, the range of $x$ for which each expansion is valid: (a) $(3 - 2x)^{-1/2}$ (b) $(1 + 5x)^{4}$.
▶ Show solution
(a) $(3 - 2x)^{-1/2} = 3^{-1/2}\left(1 - \tfrac23x\right)^{-1/2}$, valid for $\left|\tfrac23x\right| < 1$, i.e. $|x| < \tfrac32$.
(b) The power $4$ is a positive integer, so the expansion has only five terms and is exact: valid for all $x$.
Expand $\sqrt[3]{8 - x}$ up to $x^2$. Use your answer with a suitable value of $x$ to estimate $\sqrt[3]{7.9}$ to $6$ decimal places.
▶ Show solution
$(8 - x)^{1/3} = 8^{1/3}\left(1 - \tfrac{x}{8}\right)^{1/3} = 2\left[1 + \tfrac13\left(-\tfrac{x}{8}\right) + \dfrac{\frac13\left(-\frac23\right)}{2}\left(-\tfrac{x}{8}\right)^2\right]$
$= 2\left[1 - \tfrac{x}{24} - \tfrac{x^2}{576}\right] = 2 - \tfrac{x}{12} - \tfrac{x^2}{288} + \cdots$, valid for $|x| < 8$.
$x = 0.1$: $2 - 0.0083333 - 0.0000347 = 1.991632$ (6 d.p.).
(Calculator: $1.9916317\ldots$ — correct to $6$ d.p.)
In the expansion of $(1 + kx)^{-1/2}$, the coefficient of $x^2$ is $24$ and the coefficient of $x$ is positive. Find $k$, the coefficient of $x$, and the range of validity.
▶ Show solution
Coefficient of $x^2$: $\dfrac{\left(-\frac12\right)\left(-\frac32\right)}{2}k^2 = \tfrac38k^2 = 24$, so $k^2 = 64$ and $k = \pm 8$.
Coefficient of $x$: $-\tfrac12k$. This is positive only if $k = -8$, giving coefficient $4$.
$(1 - 8x)^{-1/2}$ is valid for $|8x| < 1$, i.e. $|x| < \tfrac18$.
Show that $\sqrt{\dfrac{1 + x}{1 - x}} \approx 1 + x + \tfrac12x^2$ for small $x$. By substituting $x = 0.1$, find an approximation for $\sqrt{11}$.
▶ Show solution
$(1 + x)^{1/2} = 1 + \tfrac12x - \tfrac18x^2 + \cdots$ and $(1 - x)^{-1/2} = 1 + \tfrac12x + \tfrac38x^2 + \cdots$
Product: $1 + \tfrac12x + \tfrac38x^2 + \tfrac12x + \tfrac14x^2 - \tfrac18x^2 = 1 + x + \tfrac12x^2$ ✓ (valid for $|x| < 1$).
$x = 0.1$: $\sqrt{\dfrac{1.1}{0.9}} = \sqrt{\dfrac{11}{9}} = \dfrac{\sqrt{11}}{3} \approx 1 + 0.1 + 0.005 = 1.105$.
$\sqrt{11} \approx 3.315$ (true value $3.3166\ldots$).
Expand $(1 + x)(1 - x)^{-1}$ up to $x^3$, and hence write down the coefficient of $x^r$ for $r \geqslant 1$.
▶ Show solution
$(1 - x)^{-1} = 1 + x + x^2 + x^3 + \cdots$
$(1 + x)(1 + x + x^2 + x^3) = 1 + 2x + 2x^2 + 2x^3 + \cdots$, valid for $|x| < 1$.
Every power $x^r$ with $r \geqslant 1$ arises from $1 \cdot x^r + x \cdot x^{r-1}$, so its coefficient is $2$.
(a) Express $\dfrac{1 + 5x}{(1 - x)(1 + 2x)}$ in partial fractions. (b) Hence find its expansion up to $x^3$, and state the range of validity.
▶ Show solution
(a) $1 + 5x \equiv A(1 + 2x) + B(1 - x)$.
$x = 1$: $6 = 3A$, $A = 2$. $x = -\tfrac12$: $-\tfrac32 = \tfrac32B$, $B = -1$.
$\dfrac{2}{1 - x} - \dfrac{1}{1 + 2x}$
(b) $2(1 - x)^{-1} = 2 + 2x + 2x^2 + 2x^3$ for $|x| < 1$.
$(1 + 2x)^{-1} = 1 - 2x + 4x^2 - 8x^3$ for $|x| < \tfrac12$.
Subtract: $1 + 4x - 2x^2 + 10x^3 + \cdots$, valid for $|x| < \tfrac12$.
(a) Show that $\dfrac{1}{\sqrt{4 - x}} = \dfrac12 + \dfrac{x}{16} + \dfrac{3x^2}{256} + \cdots$ and state the range of validity.
(b) Hence show that $\dfrac{2 + x}{\sqrt{4 - x}} \approx 1 + \dfrac58x + \dfrac{11}{128}x^2$.
(c) By substituting $x = 0.04$, show that $\dfrac{3.4}{\sqrt{11}} \approx 1.0251375$, and hence find an approximation for $\sqrt{11}$, giving your answer to $6$ decimal places.
▶ Show solution
(a) $(4 - x)^{-1/2} = 4^{-1/2}\left(1 - \tfrac{x}{4}\right)^{-1/2} = \tfrac12\left[1 + \left(-\tfrac12\right)\left(-\tfrac{x}{4}\right) + \dfrac{\left(-\frac12\right)\left(-\frac32\right)}{2}\left(-\tfrac{x}{4}\right)^2\right]$
$= \tfrac12\left[1 + \tfrac{x}{8} + \tfrac38\cdot\tfrac{x^2}{16}\right] = \tfrac12 + \tfrac{x}{16} + \tfrac{3x^2}{256}$ ✓, valid for $|x| < 4$.
(b) $(2 + x)\left(\tfrac12 + \tfrac{x}{16} + \tfrac{3x^2}{256}\right) = 1 + \tfrac{x}{8} + \tfrac{3x^2}{128} + \tfrac{x}{2} + \tfrac{x^2}{16} + \cdots$
$= 1 + \tfrac58x + \left(\tfrac{3}{128} + \tfrac{8}{128}\right)x^2 = 1 + \tfrac58x + \tfrac{11}{128}x^2$ ✓
(c) $x = 0.04$ is within $|x| < 4$. LHS $= \dfrac{2.04}{\sqrt{3.96}}$, and $\sqrt{3.96} = \dfrac{\sqrt{396}}{10} = \dfrac{6\sqrt{11}}{10}$, so LHS $= \dfrac{2.04 \times 10}{6\sqrt{11}} = \dfrac{3.4}{\sqrt{11}}$.
RHS $= 1 + 0.025 + \dfrac{11 \times 0.0016}{128} = 1 + 0.025 + 0.0001375 = 1.0251375$ ✓
$\sqrt{11} \approx \dfrac{3.4}{1.0251375} = 3.316628$ (6 d.p.). The true value is $3.316625\ldots$, so the error is only about $3 \times 10^{-6}$.